P_A = (235y - 125xy) , text{mm of Hg}. P_A is | vedclass

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(C) According to Raoult's law,the partial pressure of $A$ is given by $P_A = P^o_A \cdot x_A$,where $x_A$ is the mole fraction of $A$ in the liquid ph

Vedclass · Accepted Answer

$110$

Vedclass · Answer

(C) According to Raoult's law,the partial pressure of $A$ is given by $P_A = P^o_A \cdot x_A$,where $x_A$ is the mole fraction of $A$ in the liquid phase.Since $x$ is the mole fraction of $B$ in the liquid phase,$x_A = 1 - x$.Thus,$P_A = P^o_A(1 - x) = P^o_A - P^o_A \cdot x$.Also,from Dalton's law,$P_A = y \cdot P_{total}$,where $y$ is the mole fraction of $A$ in the vapour phase.Given the equation $P_A = 235y - 125xy$,we can rewrite this as $P_A = y(235 - 125x)$.Comparing the expressions,we note that for a binary mixture,$P_{total} = P^o_A x_A + P^o_B x_B = P^o_A(1-x) + P^o_B x = P^o_A + (P^o_B - P^o_A)x$.Substituting this into $P_A = y \cdot P_{total}$,we get $P^o_A(1-x) = y(P^o_A + (P^o_B - P^o_A)x)$.Rearranging to match the form $P_A = 235y - 125xy$,we identify $P^o_A = 235$ and $P^o_A - P^o_B = 125$.Therefore,$P^o_B = P^o_A - 125 = 235 - 125 = 110 \, 	ext{mm of Hg}$.

$P_A = (235y - 125xy) \, \text{mm of Hg}$. $P_A$ is the partial pressure of $A$,$x$ is the mole fraction of $B$ in the liquid phase in the mixture of two liquids $A$ and $B$,and $y$ is the mole fraction of $A$ in the vapour phase. Then $P^o_B$ in $\text{mm of Hg}$ is:

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